放缩法

放缩法是通过舍去或添加一些项来构造不等式的一种方法。^{[参 1]}

例子

求证 ${\sqrt {\log _{2}3}}+{\sqrt {\log _{3}2}}<{\sqrt {2}}+1$ ^{[注 1]}
${\sqrt {\log _{2}3}}+{\sqrt {\log _{3}2}}<{\sqrt {\log _{2}4}}+{\sqrt {\log _{3}3}}={\sqrt {2}}+1$ ^{[参 2]}
已知a,b,c,d为正数，求证 $1<{\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}<2$
${\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}>{\frac {a}{a+b+c+d}}+{\frac {b}{a+b+c+d}}+{\frac {c}{a+b+c+d}}+{\frac {d}{a+b+c+d}}=1$

${\frac {a}{a+b+d}}+{\frac {b}{a+b+c}}+{\frac {c}{b+c+d}}+{\frac {d}{a+c+d}}<{\frac {a}{a+b}}+{\frac {b}{a+b}}+{\frac {c}{c+d}}+{\frac {d}{c+d}}=2$ ^{[参 3]}
求证 $\sum _{k=1}^{n}{\frac {1}{k^{2}}}<2-{\frac {1}{n}}$
$\sum _{k=1}^{n}{\frac {1}{k^{2}}}<1+\sum _{k=2}^{n}{\frac {1}{k(k-1)}}=1+\sum _{k=2}^{n}{\frac {1}{k-1}}-{\frac {1}{k}}=2-{\frac {1}{n}}$ ^{[注 2]} ^{[参 2]}
设 $n\in N^{+}$ ，求证 ${\frac {n(n+1)}{2}}<\sum _{k=1}^{n}{\sqrt {k(k+1)}}<{\frac {(n+1)^{2}}{2}}$
$k={\sqrt {k^{2}}}<{\sqrt {k(k+1)}}<{\sqrt {k^{2}+k+{\frac {1}{4}}}}=k+{\frac {1}{2}}$

${\frac {n(n+1)}{2}}=\sum _{k=1}^{n}k<\sum _{k=1}^{n}{\sqrt {k(k+1)}}<\sum _{k=1}^{n}(k+{\frac {1}{2}})={\frac {n^{2}+2n}{2}}<{\frac {(n+1)^{2}}{2}}$ ^{[注 3]} ^{[参 3]}
设a,b,c为直角三角形的三边,c为斜边，求证： $a^{n}+b^{n}<c^{n}(n>2)$
$a^{n}+b^{n}=a^{2}a^{n-2}+b^{2}b^{n-2}<a^{2}c^{n-2}+b^{2}c^{n-2}=c^{n}(n>2)$ ^{[注 4]} ^{[参 2]}

^ 用放缩法证明不等式. （原始内容存档于2014-07-26）.
^ ^2.0 ^2.1 ^2.2 董卫平. 说说放缩法. 数学大世界(高中). 2011, (2) [2015-09-20]. （原始内容存档于2016-03-04）.
^ ^3.0 ^3.1 例谈不等式证明的十种常用方法. [2014-07-16]. （原始内容存档于2014-07-20）.